A Study of Operating Characteristics of Pump-as-Turbine (PAT) for Small Scale Power Generation – A

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View with images and charts A Study of Operating Characteristics of Pump-as-Turbine (PAT) for Small Scale Power Generation – A Simulation Approach Chapter 1 Project Overview 1.1 Introduction Energy is one of the most fundamental elements of our universe. Human being use energy to do work, lights our cities, powers our vehicles, trains, planes and rockets, warms our homes, cooks our food, plays our music and gives us pictures on television. We have no other alternative for our survival. Bangladesh is facing serious energy crisis problems in rural areas including suburban and coastal region, especially in the hilly areas. Now days, it has become a burning issue of Bangladesh. Lack of proper maintenance of the machines and equipments, many years old setup in production of electricity have made the situation even worse. The demand for electricity is increasing day by day for burgeoning population as well as industries. These days, fossil fuel based energy systems are well established and far superior than renewable energy technology, people are still thinking of heading towards the renewable energy technology. The main reason of shifting to a new technology is mainly due to the fact that this earth holds no more fossil fuels. Still researchers are searching of extracting sites of fossil fuels but it is estimated that this fossil fuels that we have may last for next 50-70 years or perhaps 100 years at max. For the case of Bangladesh, there is another big problem. It is the lack of maintenance of the power generating plants. Due to the lack of maintenance, now we see that we have power plants of huge capacity but they are unable to generate any power. Another big factor comes out too, ie, people are now more conscious about the environment. Burning fossil fuels produces green house gas like CO 2, This gas is responsible for the global warming of the planet and for that reason now we see that the Polar Regions are melting and the sea level is rising causing a threat to the low lying countries. Another factor is the Acid Rain. Acid rains cause damage to crops, buildings, aquatic lives forests and human health. On, the other hand, renewable sources are not dirty and they are not harmful to our


environment. So, this is our very time that we move away from the fossil fuel based technology and embrace the renewable way of generating power. For, this we can use the small scale generation of using electricity PAT. It is cheap, safe and of course not harmful to the environment. The people of urban areas usually distribute water by pipe line. A PAT can be set up across these pipes. This type of generation may be a potential energy source for urban areas in Bangladesh. This thesis is designed for generating power in those areas where there is water at a certain level above the ground, ie, fall or springs and using water’s potential energy to convert to our electrical energy. We know, Potential Energy = mass x acceleration due to gravity x height. Let, mass be m acceleration due to gravity be g height be h So, Potential Energy of the water = mgh………………………………………………………. (1) Here mass m can be written as Flow x Density = Q x ρ ……………………………………………………………………………….. (2) Therefore, Potential Energy = Flow x Density x acceleration due to gravity x height. = ρQgh…………………………………………………………………….……….(3) So, this is the energy available in the water. If we can harness some of its energy, we get power.


To achieve the power using the motor-pump, first of all we need to know what a pump is and what a motor is. Also some characteristics of the pump and the motor are needed to be known. 1.2 Project Technique What does a motor-pump do? A motor-pump is a machine that is connected by a shaft with a motor. The motor takes in electrical energy and starts to rotate. So, the electrical energy is first converted to the kinetic energy of the motor. The shaft also rotates. The other end of the shaft is connected to the pump which also rotates. There is a liquid on which the pump will work, say it is water. The pump imparts this kinetic energy to the liquid. This kinetic energy is largely transformed into the pressure energy as the liquid leaves the pump. Here is Figure that shows the principle of a motor-pump.

Figure 1.1 Flow diagram of a motor-pump

Our aim is to do the opposite work. Now water is available at a certain height. So, this water has potential energy. The water is allowed to flow through the pump at a certain rate. This converts the potential energy of the water into the mechanical energy of the pump-as-turbine. An induction motor is attached to the pump-as-turbine. The shaft of the motor will rotate and induce voltage in the stator of the motor so that the motor is now working as generator. Here is Figure that shows the principle of PAT.

Figure 1.2 Flow diagram of a PAT


Hence the main objective of this thesis is to get the idea if it is possible to use a motor-pump as a generating mechanism for electricity. In doing so, the operational characteristics of the pump in the turbine mode, ie, Pump-as-Turbine (PAT) has to be known. But the PAT mode characteristic is not easy to obtain as the pump manufacturers do not provide them. They only give the pump mode characteristics. So, the PAT characteristic is obtained following the procedures of two scientific researchers Sharma and Chapallaz et al. Hypothetical site curves are also drawn and the operating point of the PAT is obtained from where the output power can be found. The value of the magnetizing capacitance is also found. 1.3 Advantages and disadvantages of motor-pump 1.3.1 Advantages A motor-pump has the following advantages: •

Availability Due to its wide spread application, a motor-pump is readily available anywhere in the market. If spare parts are needed, incase, they are equally found in the market as the motor-pump manufacturers often provide after-sales services almost throughout the world.

Economic Advantage The total cost of a motor-pump is very less compared to that of buying separate synchronous generator and turbine. A pump may cost as much as 50% less than that of a turbine.

Construction Standard motor-pumps are very simple and robust in construction. They do not require any highly qualified mechanics for maintenance. The rotors have no windings, diodes or slip rings. They are also designed for continuous operation.

Environment Friendly


PAT is very much environment friendly. It does not burn any sorts of fuel. So, there is no chance of emission of harmful gases that readily harms the environment. It keeps the air and its surrounding clean and green. •

Per Unit Charge PAT is very promising about the per unit charge. It does not require any sorts of fuel. So, the cost of electricity that it is producing is constant. Moreover, the per unit charge is very less as compared to the Electricity distributors as it produces electricity from a renewable source. Also, it helps the country’s economy by saving lots of fuel from burning.

Easy Installations The parts and arrangements required by the PAT are very simple and readily available. Many local technicians can easily setup such kind of system.

1.3.2 Disadvantages Every machine has some drawbacks. The followings are the disadvantages of the PAT. •

Site Dependent The most important and significant drawback of the PAT is its site dependency. It cannot be built on any locations. It requires particular infrastructures where it can be used. It requires a constant head and flow rate for its operation.

Hydraulic Control Device The PAT requires some control valves incorporated to the penstock line for its operation. This adds additional costs to the system.

Calculations Required A PAT requires some calculations for generating the power. An induction machine connected with the pump-as-turbine will not work unless suitable value of the capacitor is used.

1.4 Applications


Applications range from direct drive of machinery in agro-processing factories and small industries (flour mills, oil expellers, rice hullers, saw mills, wood and metal workshops) to electricity generation both in stand-alone and grid-linked stations.

A common feature of these installations is the need to reduce pressure of a fluid at the end of a process or before further application or treatment of the fluid. The conventional approach would dissipate this surplus pressure in any kind of throttling device (valve, orifice, stilling basin). Using a reverse running pump as a hydraulic power recovery turbine (HPRT) recovers this energy at almost no costs thereby improving overall plant efficiency.

Chapter 2 Pump and its characteristics 2.1 Introduction A pump is a machine that displaces liquid from one position to another position. It works on a liquid. It converts its kinetic energy (obtained from a motor) to the potential energy of the liquid. There are varieties of pumps available. They are: •

Centrifugal pump

Axial flow pump

Submersible pump

Positive displacement pump

Self-priming pump

Of the above mentioned pumps, Centrifugal pump is the most economical one and they are most widely used. Its maintenance cost is very low and is readily available in the market. It comes with an induction motor coupled with it. 2.2 Centrifugal Pump


It uses the centrifugal force to pump a liquid. It takes in water axially and discharges it with the help of centrifugal force. A centrifugal pump is a very simple machine. It has two main components. They are the impeller and the diffuser. 2.2.1 Impeller Impeller is the moving part of the pump. It is attached to a shaft and driven by a motor. Impellers are generally made of bronze and stainless steel. 2.2.2 Diffuser It is also called Volute. It houses the impeller. It captures the liquid and directs the liquid off the impeller.

Figure 2.1Centrifugal Pump with impeller and diffuser

2.2.3 Centrifugal Pump Operation The water enters the centre of the impeller and exits the impeller with the help of the centrifugal force. As it leaves, a low pressure is created causing more water to flow into the centre. The high speed spinning of the impeller causes velocity to be developed to the water. The water velocity is collected by the diffuser and converted into pressure. The pressure is the head that is developed by the pump.


2.3 Pump Curves Any machine’s performance can be expressed graphically by some means of curves. For a centrifugal pump there are certain curves that tell us about the performance of the pump and also their bep and operating points. Such curves are illustrated below: 2.3.1 Head-Flow Curve It is one of the crucial characteristics of a pump. It shows the variation of the Head with the change in flow rate, Q. According to the Figure 2.2, when the flow is zero, the head is at fixed value called the ‘shut-off’ head. As the flow increases, the head decreases. Suppose the pump moves the water from a lower level to an upper level. Initially, at the discharging point, there is no water. Now, as the discharging of water takes place, the height of the water at the discharge point increases. So, the pressure at the discharge point also increases. This causes a decrease in the head as the flow increases. Obtaining the Head-Flow Curve Initially, the pump is turned on. The delivery valve is completely closed. The flow and the manometric pressures at the delivery and at the suction are noted. Now, open the delivery valve a little, and take the above readings. The experiment is repeated for several positions of the valve and the corresponding pressures and the flow rates are measured. The following equations give the values of head and the flow rates: Hm = (P2-P1)/ ρg + (V22-V12)/2g + (Z2-Z1)…………………………………………………………(4) V2 =Q/A2………………………………………………………………………………………………..(5) V1 =Q/A1………………………………………………………………………………………………………………………..…………………….....(6) 2.3.2 Efficiency Curve The Q vs pump efficiency of the pump is an inverted ‘U’ shaped curve. At no flow, The efficiency is zero. It then rises to a maximum flow rate at the bep point. Beyond this point, the flow starts decreasing. The pump is designed to work at the bep point but due to different flow rates, it cannot operate on such point.


Obtaining the Efficiency Curve Initially, the pump is turned on. The delivery valve is completely closed. The flow and the manometric pressures at the delivery and at the suction and the input voltage and the current are noted. Now, open the delivery valve a little bit, and take the above readings again. The experiment is repeated for several positions of the valve and the corresponding pressures, flow rates, currents and voltages are measured. Hm is found using the equations (4). The hydraulic power, Ph = ρgQHm…………………………………………….………………..… (7) The electric power, Pe = VI…………………………………………………………………………. (8) The overall efficiency of the pump, ɳo = Ph/Pe …………………………………………………....(9)

Figure 2.2 Head-Flow Curve

Figure 2.3 Efficiency Curve


Chapter 3 Pump-as-Turbine (PAT) 3.1 Introduction A turbine is a rotary engine that extracts energy from a fluid flow and converts it into useful work. Rotational fluid machines are completely reversible and a pump can run effective as a turbine. Reversing the direction of the flow through the same impeller will make the pump work as turbine. 3.2 Difference between Pump and Turbine The basic hydraulic theory is the same for both machines. However, the behavior of real fluid flow including friction and turbulence results in different rules for the design of pumps and turbines.

Figure 3.1 Main difference of fluid and energy flow in pumps and turbines


Some particular differences a) Operating conditions: PUMPS are usually operated with constant speed, head and flow. A pump is therefore designed for one particular of operation (duty point) and does not require a regulating device (guide vane). Ideally, the duty point coincides with the maximum efficiency of the pump. TURBINES operate under variable head and flow conditions. In a PAT, flow must be adjustable to either accommodate to seasonal variations of the available water or to adjust power output according to the demand of the consumers. Adjustable guide vanes and/or runner blades (or nozzles controlled by a streamlined valve) regulate the flow (see Figure 3.2). b) Hydraulic design: In a PUMP, kinetic energy imparted to the fluid must be converted into pressure energy; i.e. flow must be decelerated as it passes through the impeller and the volute casing. Decelerated flow is generally very sensitive to separation and the formation of eddies. To avoid these, impeller passages are made of long smooth channels with gradually increasing cross-sectional area. Friction losses through these long passages are relatively high (see Figure 3.2). On the other hand, flow through a TURBINE is accelerated which is less subjected to turbulence; runner passages are therefore relatively short which reduces friction losses and ensures high efficiencies c) Location of the machine – Cavitation


The physical location of the PUMP in relation to the water level of the sump from which water is being pumped is critical. If it is too high, cavitation may occur. In this context, suction pipe design is very important since friction losses in the suction pipe reduce pressure at the pump inlet and increase the tendency to cavitation. TURBINES are less sensitive to cavitation since friction losses in the draft tube increase the backpressure on the turbine.

Figure 3.2 Comparison of flow conditions in a pump impeller and a turbine runner

The behavior of real fluid flow including friction and turbulence results in different rules for the design of pumps and turbines. 3.3 Pump-as-Turbine Head and Flow


When the pump is operated in the turbine mode, ie, the direction of the flow is reversed, we get the PAT curve.

Figure 3.3 PAT head and flow

When the flow increases, the head also increases. 3.4 Site Curve The curve that tells us about the site is called the site curve. It’s a head versus flow curve. For a site, the head is always fixed. But due to head losses at higher flow rates the curve has a drooping effect The equation relating the head losses to flow and the length of the pipe is given below. Hf/Hsite (%) = (Q/Qtable)2 x Lpipe/Hsite ……………………………………………………………(10) where

hf is the head loss


Hsite is the gross head in meters Lpipe is the length of the pipe in meters Q is the flow rate through the turbine in l/s Q table is the flow rate given in Table

Figure 3.4 Site Curve

3.5 Turbine Operating Point The intersecting point of the PAT curve with the site curve is the operating point of the turbine. It is not necessary that the operating point need to be the maximum efficiency point, it may vary, but it is preferred that the operating point is close to the maximum efficiency point of the turbine.

Figure 3.5 Intersection of the PAT curve with the Site Curve gives the operating point


3.6 Methods of Analysis 3.6.1 Method by Sharma A simple method was proposed by Sharma (1985) where the flow rate and the head at the bep for the pump (Qbep and Hbep) are related to the turbine flow rate and head (Qt, Ht, and ɳt) by the maximum efficiency (ɳmax) of the pump. The following equations describe the relationship. Qt = Qbep/ɳmax0.8 x (Nt/Np) …………………………………………………………………...…….. (11) Ht = Hbep/ɳmax1.2 x (Nt/Np)^2………………………………………………………………………. (12) ɳt = ɳmax ……………………………………………………………………………………………....(13) Pout = ɳx Htx Qtx9.81x1000………………………………………………………………………..(14) Pe = 0.73x Pout……………………………………………………………………………………….(15)

The method by Sharma only provides the Head and Flow required by a pump to operate in the PAT mode in the maximum efficiency. The efficiency of the PAT in converting the mechanical power into electrical power is 73%. So, Pe is multiplied with a factor of 0.73. 3.6.2 Method by Chapallaz et al. The method proposed by Shrama would only give the point where the PAT should be operated but it does not yet give the PAT curve. Later Chapallz et al. (1992) gave another method of determining the operating point of the PAT. He not only gave the operating point but he also showed a method of determining the PAT curve. His method uses experimental data to determine empirical correlations but in this case over 80 different pumps were used. His method follows the following procedures:


First, determine the rated pump head, flow and the maximum efficiency. These values are typically found on the data sheets of most pump manufacturers. Now, calculate the specific speed of the pump using the equation below: ……………………………………………………………………... (16)

The conversion factors CH and CQ can be read from the graphs in the Figures of Appendix A using the maximum efficiency of the pump. Now, the following scatter factors will determine the performance range of the PAT. CHmax = 1.1CH ……………………………………………………………………………..(17) CHmin = 0.9CH ……………………………………………………………………………..(18) CQmax=1.075CQ ..………………………………………………………………………….(19) CQmin = 0.925CQ …………………………………………………………………………..(20) Now, it is simple to determine the maximum and the minimum turbine design head and flow at the rated pump speed and then convert this to the nominal turbine speed by substituting the appropriate factors in the following equations.

.………………………………………………………...(21)

……………….………………………………...…………....(22)

…………………………………………………………………......(23)

…………………………………………………………………........(24)

And

ɳt = ɳmax – 0.03…………………………………………………………………………………….....(25)


Maximum output Power = ɳtmax .Htmax .Qtmax .9.81 .1000 ……………………………………...(26) Minimum output Power = ɳtmax .Htmin .Qtmin .9.81 .1000 ……………………………………….(27)

So, the minimum and the maximum power output can be obtained using head, flow rate and efficiency. In order to determine the shape of the curve away from the bep, the graphs in Appendix B and Appendix C are used. These diagrams give the head and power as function of the flow rate for various specific speeds. The resulting curves can now provide a good estimate of the performance of the pump in turbine operating mode.

Chapter 4 Three Phase Induction Generator 4.1 Introduction to Induction Machine We all know of Induction motors. An induction motor is an asynchronous motor which when given AC power supply will begin to rotate, less than the synchronous speed. The power is supplied by the stator to the rotor by means of electromagnetic induction. 4.2 Construction of Induction Machine The Induction Motor consists of two parts: a fixed part, ie, the stator which is wounded by some coils. The function of the stator is to generate a rotating field. The moving part is called the rotor. It is usually made in the shape of a squirrel cage and often called as the squirrel cage rotor. It consists of bars shorted at both ends rather than conductors.

4.3 Operation of Induction Machine When an induction motor is connected to a three phase supply, a rotating field is setup in the core. The rotating magnetic field cuts the short-circuited bars. According to the Faraday’s Law of electromagnetic induction, this induces an emf. As the ends are shorted, a current is produced in the rotor. This current produces a magnetic field in the rotor that interacts with the stator field. According to the Lenz’s law, the direction of this field is such that it opposes


the very cause that producing it. So, a torque is developed and the motor starts rotating. It never reaches the synchronous speed at which the field is rotating, otherwise there will be no flux linkage and the motor will stop.

Figure 4.1 Two Three Phase Induction Motors

4.4 Slip Slip is defined as the difference between the speed of the rotating magnetic field and the rotor speed. Let, the rotating field speed be ns the rotor speed be nr then, the slip, s=( ns – nr)/ ns ……………………………………………………………………(28) So, the slip is always less than 1.


4.5 The Induction Generator

Figure 4.2 Torque-Speed Characteristics of an Induction Motor above its synchronous speed

The Figure 4.2 shows the torque-speed characteristics of an induction motor when operated above its synchronous speed. So, if the same supply-connected induction machine is operated above its synchronous speed, it will become a generator. In the generator mode, the slip is negative. But in order to create a rotating field, it still needs magnetizing current from the supply, just as though it were a motor. From the above diagram, it can be seen that the full load power can be achieved if it is operated at slip negative to the full load motoring slip. 4.6 The Induction Generator Operating Alone For the induction machine to operate alone, some magnetizing currents are needed to be supplied. This can be done using external capacitors connected to the machine because capacitors are the only external sources of magnetizing current. Therefore, in order to obtain the required operating voltage at the desired frequency, the amount of capacitance must be carefully chosen.


Figure 4.3 shows the variation of the magnetizing current with the terminal voltage of an induction motor. This can be obtained by running the machine as a motor at no load and measuring its armature current as a function of terminal voltage. To achieve a given voltage level in an induction generator, external capacitors must supply the corresponding magnetization current corresponding to that level.

Figure 4.3 Variation of the magnetizing current with

Since the reactive current that a capacitor can produce is directly proportional to the voltage applied to it, the locus of all possible combinations of voltage and current through the capacitor is a straight line. Such a plot voltage versus current for a given frequency is shown in the Figure 4.4.

Figure 4.4 I-V Characteristics of Capacitor


Plotting both the graphs at the same scale gives the operating points of the induction generator as shown in Figure 4.5.

Figure 4.5 Operating Point of the generator

4.6 Voltage Build up in Induction Generator An induction generator can only be turned on if there is sufficient residual magnetism in the rotor. Residual is the initial magnetism present in the rotor steel. This magnetism in the field circuit produces a small voltage. That small voltage produces a capacitive current flow, which increases the voltage, further increasing the capacitive current and so forth until the voltage is fully build up. If there is insufficient residual magnetism, the generator speed can be increased because at higher frequency, less magnetism is required for the excitation to occur. However, if this still does not work, a dc supply for a few seconds across any two machine terminals can be used for the excitation to occur. A car battery is more than sufficient for this purpose. 4.7 Excitation Capacitor Requirements 4.7.1 No load test


It is used to calculate the excitation capacitance. This is because the apparent power drawn by the motor at no load is approximately the same reactive power required while running as generator close to full load. This is true for machines below 5kW.

4.7.2 Capacitances in Star and Delta Capacitors has to be connected either star or delta for three phase generating system. Figure 4.6 shows how they can be connected as star or delta.

Figure 4.6 Star and Delta Connections of excitation capacitance

The relation between the star and the delta connection are related as follows:

………………………………………………………..………... (29) ………………………………………………………………........(30)

…………………………………………………………………....(31)

………………………………………... (32)


Therefore,

…………………………………………………………………....(33)

So, it can be concluded that if capacitors are connected in star, then one-third of capacitance is required for delta connection. 4.7.3 C-2C method One way to get a single-phase output from a three-phase generator is to use the C-2C method. Figure 4.7 shows the circuit diagram.

Figure 4.7 Single phase output from a three phase system

The method for obtaining a single-phase output from a three-phase machine is as follows: •

Use a three-phase induction machine suitable for 240/415 V operation and connect the machine in delta.

Calculate the per phase capacitance (calculation has been done in topic 4.7.4 Calculation), ‘C’, required for normal three-phase 240V delta operation.

Instead of connecting ‘C’ to each phase connect twice ‘C’ to one phase, ‘C’ to a second phase and no capacitance to the third. The load is placed across the ‘C’ phase.


This unbalanced arrangement of capacitance helps to compensate for the unbalanced load on the generator, and as a result the generator can be used with an output of 80% of the motor rating. While using the ‘C-2C’ connection, it is essential to ensure that the direction of rotation of the machine rotor is correct, in relation to the phase to which ‘C’ and ‘2C’ values of capacitance are connected. If capacitors are arranged correctly the ‘C’ phase will produce its peak voltage before the ‘2C phase’. If opposite occurs the generator will run inefficiently and overheat.

4.7.4 Calculations Considering a three phase motor attached to a 4 horse power pump. The motor draws I current at V voltage. Therefore, Total apparent power drawn by the motor at no load is = = S VA ……………………………………………………………………..(34) At no load, the current drawn by the motor is approximately the current due to the reactive power. Therefore, Q = S var………………………………………..…………………………(35) Qp (per phase) Vp (per phase) =V/

= S/3 ………………………………………………...……………………..(36) …………………………………………………………………….…(37)

Ip (per phase)

=Qp/Vp …………………………………………………………………….(38)

Capacitace

= Ip/(2 π fVp) …………………………………………………....…………(39)


4.8 Operating Voltage and Frequency The operating voltage and frequency will depend on the followings: •

Turbine Power output.

Excitation Capacitance.

Load and power factor.

4.8.1 Operating Voltage We have seen earlier that for constant speed, the operating voltage is determined by the amount of connected. For operation at synchronous speed, the highest voltage that can be achieved is typically 125% of the motor rating and the lowest is approximately 65% of the motor rating. The upper limit is set by the voltage at which the excitation current equals the rated current of the machine and the lower limit is the voltage at which the machine is sufficiently saturated to operate stably. Operation at close to the upper voltage limit is not recommended, because, due to winding temperature considerations, the large current flowing into the generator from the excitation capacitors reduces the maximum load that can be connected and reduces the efficiency of the rotor. Operation at close to the lower voltage limit is not recommended for two reasons: •

The reduced excitation capacitance makes the generator very sensitive to reactive loads, which could cause significant variations is frequency and possibly loss of excitation.

The efficiency will be below the optimum efficiency, particularly when the generator is heavily loaded. This is because, in order to achieve the same power output at reduced voltage, the load current must be increased. This increases rotor and stator heating and reduces efficiency.

Also when selecting the operating voltage, the transmission line voltage drop must be considered.


4.8.2 Operating Frequency Increasing the operating frequency will reduce the excitation current required to achieve the rated voltage. This improves efficiency and maximum power output. 4.9 Effect of load on Generator Output 4.9.1 Resistive Load Figure 3.8 below shows the variation of resistive load upon induction generator operating at constant speed and fixed excitation capacitance. The full load current is the rated motor current. If load exceeds, the excitation will collapse. Also Figure 3.9 shows the turbine power variations with speed. With high loads, the speed of the turbine will decrease and so as the frequency. This automatically decreases the voltage drop and reduces excitation capacitance available. The turbine speed and the generated voltage will fall until a speed is reached at which the power output of the turbine equals the load on the turbine. Alternatively, it can be said that if the resistive load is decreased, the generator voltage will increase and the turbine speed too.

Figure 4.8 Variation of phase voltage with resistive load

Figure 4.9 Turbine Power variation with Speed


It is thus clear that the resistive load must be controlled to maintain the voltage and frequency. 4.9.2 Inductive Load For an inductive load connected to the induction generator, the current required by the load will be supplied from the excitation capacitances. This will reduce the amount of excitation available, and the voltage will drop temporarily. The drop in voltage will reduce the power of the other loads connected to the generator. So the turbine speed will increase. The voltage will increase until the power output of the turbine equals the power required by the load. Hence inductive load increments result in an increased frequency. If a very large inductive load is connected, the excitation of the generator will collapse. 4.9.3 Capacitive Load For a capacitive load connected to the induction generator, the current required by the load will be supplied from the excitation capacitances. This will increase the amount of excitation available, and the voltage will increase. So the turbine speed will increase. The voltage will increase until the power output of the turbine equals the power required by the load. Hence capacitive load increments result in an increased frequency. 4.10 Generator Protection The generator must be protected under any circumstances. This topic deals with the generator protection under Overload, Underload and Overspeed conditions. 4.10.1 Overload


The overload on an induction generator will not be possible as the excitation will collapse due to excess load and will limit the winding currents to safe values. However, it is possible that one of the phase from the three phase connection to burn out if it is severely overloaded. So, Miniature Circuit Breakers (MCBs) connected to each phase could be used rated 1.5 times the rated full load generator output. 4.10.2 Underload An induction generator is at a much higher risk at underload conditions rather than at overload conditions. This is because, when the load is less, the speed of the generator will increase so as the frequency and the voltage. Both this frequency and the voltage will increase the excitation current flowing into the generator and may burn out the windings. So, MCBs should be connected in series with the capacitances so as to trip when the speed is high. 4.10.3 Overspeed Standard induction machines can withstand considerable overspeeds due to their solid rotors. Twice the rated speed is normally acceptable for 2 pole machines.

Chapter 5 Design Criteria of the Civil Work 5.1 Introduction This chapter will be dealing with the necessary information for the civil works of the PAT. 5.2 Intake Requirements Intake is the structure from where the water is brought to the PAT scheme. Its main purpose is to divert the water from the fall or stream to the reservoir. It will also separate the sand, stones, silt and floating particles from the water preventing the blockage of the PAT. 5.3 Reservoir A PAT system will generate electricity from water running down from falls or streams. First this water has to be collected and reserved in a certain area called the reservoir. It has to be constructed just beside the fall or the stream to maintain the height. The reservoir will hold


the water up to a certain height. Over that height, it will discharge the water from another outlet. The structure can be made out of bricks, rods and cements.

5.4 Penstock The penstock is the pipe that leads the water to the turbine. It is the most expensive part in the PAT scheme. So, it must be made sure that the cost of the penstock is as small as possible. The choice of the pipe diameter will determine the head loss and will depend on the expected flow rate. Since the PAT is a fixed flow unit, the penstock diameter can be chosen to match the required flow and the head loss calculated accurately. Appendix D shows how to select the most suitable size of the penstock and how to calculate the head loss. 5.5 Valves A valve will be needed to maintain the required flow rate. These valves are often found in the market readily. It is usual practice to place the valve at the end of the penstock before the water enters the PAT. Figure 5.2 below shows a diagram of a valve placed just above the PAT.

Figure 5.2 Gate Valve and Outlet Channel


5.6 Outlet Channel At the outlet from the PAT, it is advisable to use an expanding pipe know as a draft tube, or a reducer in reverse to increase the diameter of the pipe where it discharges into the outlet channel. The water in the outlet channel is at the atmospheric pressure and therefore the channel must be large enough and with a great enough slope to take away the full flow from the PAT.

Figure 5.3 Schematic Diagram of PICOO-HYDRO


Chapter 6 Experimental Data and Calculations 6.1 Introduction This chapter deals with the data of a pump which was conducted experimentally. From the experiments several data were taken to plot the pump characteristic curves, ie, the Head-Flow Curve of a pump and the Efficiency Curve of a pump. The bep of the pump were found. Here is a diagram of the Pump on which the experiment was carried out.

Figure 6.1 4 horse-power DC powered Motor Pump from IUT

6.2 Data Recorded The following table shows the data that have been taken from the above Figured machine. Six sets of data were taken while carrying out the experiment. Table 1 Data Recorded for the Pump Characteristics

Q[m^3/hr]

Pinlet[bar]

Poutlet[bar]

V[v]

I[A]


23

-0.16

2.4

143

21.6

19

-0.12

2.8

143

19.2

15

-0.08

3.1

143

17

11

-0.06

3.25

143

14.6

7

-0.04

3.4

146

11.8

3

-0.03

3.4

146

9.4

6.3 Modified Data Table The readings taken were not sufficient to draw the required Head-Flow Curve and the Efficiency Curve. Some calculations were made and a modified table was made. This table includes more relevant data that are needed to draw the above mentioned curves. Equations (4), (5), (6), (7), (8) and (9) were used to find out the followings: V1[m^3/s], V2[m^3/s], Hm[m], Ph[W], Pe[W] and Efficiency. Table 2 Data Adjusted for Pump Characteristics

Q[m^3/s]

V1[m^3/s] V2[m^3/s]

P1[Pa]

P2[Pa]

Hm[m]

Ph[W]

Pe[W]

ɳ

0.006389

3.15189

5.59937

-16000

24000 0

27.43 7

1719.6 4

3088.8

0.55

0.005278

2.6037

4.6255

-12000

28000 0

30.76 0

1592.6 2

2745.6

0.58

0.004167

2.05558

3.65176

-8000

31000 0

33.13 0

1354.1 9

2431

0.55

0.003056

1.50742

2.67796

-6000

32500 0

34.24 0

1026.3 6

2087.8

0.49

0.001944

0.9592

1.70415

-4000

34000 0

35.41 7

675.58

1722.8

0.39

0.000833

0.41111

0.73035

-3000

34000 0

35.23 2

288.02

1372.4

0.21

6.4 Curves Since we have few data readings of the head, flow and efficiency the Head-Flow Curve and the Efficiency Curve can be drawn. From the respective curves drawn in the same graph, the bep point of the pump can be found. The following Figures show the curves.

Figure 6.2 Head-Flow Curve


Figure 6.3 Efficiency Curve

Plotting both the curve in the same graph will give the best efficiency point (bep) of the pump.

Figure 6.4 Head-Flow Curve and Efficiency Curve with the Best Efficiency Point


From the above diagram, it can be noted that the bep of the pump is 19 m3/hr and 30.2 m and ɳp is 0.58.

Chapter 7 Findings and Analysis 7.1 Introduction This chapter deals with the calculations of the methods approached by Sharma and Chapallaz. Et al. 7.2 Method by Sharma The pump was operated at Np 2900 rpm. But the synchronous speed of the pump was 3000 rpm. Using equation (24), s = 0.033. So, the slip at which the motor should operate is -0.033. Therefore, the speed at which the turbine should rotate is Nt 3100 rpm. Using equation (11), Qt = 31.4 m3/hr


Using equation (12), Ht = 66.3 m Using equation (13), ɳt = 0.58 Using equation (14), Pout = 3290 W Using equation (15), Pe = 2401 W A 4 horse-power machine with 3 phase induction motor requires 3088.8 W of input power. With the machine working as PAT, it will deliver 2401 W. 7.2.2 Capacitance Requirements Considering a three phase motor attached to a 4 horse power pump. The motor draws approximately 7.2 A of currents at 415 V. Therefore, Using equation (34), Apparent Power

= 1.73x415x7.2 = 5176 VA

Therefore, Using equation (35), Reactive Power Q

= 5176 var.

Using equation (36), Qp (per phase) = 5176/3 = 1725 var Using equation (37),


Vp (per phase) = 415/1.73 = 240 Using equation (38), I p(per phase) =1725/240 = 7.2 A Using equation (39), Capacitance = 95.5x 10-6 F. 7.3 Method by Chapallaz et al. Using equation (16), Np

= 16.4

Using equation (17), CHmax = 2.27 Using equation (18), CHmin

= 1.85

Using equation (19), CQmax =1.77 Using equation (20), CQmin = 1.53 Using equation (21), Htmax(nt) = 78.3m Using equation (22), Htmin(nt) = 63.8m Using equation (23), Qtmax(nt) = 35.9m3/hr


Using equation (24), Qtmin(nt) = 31.7 m3/hr Using equation (25), ɳt

= 0.55

Using equation (26), Maximum output Power

= 4213 W

Using equation (27), Minimum output Power

= 3030W

Maximum electrical output Power = 0.73*3944 = 3075W Minimum electrical output Power = 0.73*2836 = 2212W From the Appendix B and Appendix C, the followings are noted. Table 3 Factors for Calculating Head, Flow and Power away from BEP

Qt/Qnt

Ht/Hnt

Pt/Pnt

0.8

0.73

0.57

0.9

0.83

0.73

1

1

1

1.1

1.17

1.25

1.2

1.33

1.5

From the above Table Table 4 Actual Values for Head and Flow for Maximum and Minimum Deviations

Ht [maximum deviation]/m

Ht [minimum deviation]/m

Qt [maximum deviation]/m^3/hr

Qt [minimum deviation] /m^3/hr

53.509

43.581

28.72

25.36

60.839

49.551

32.31

28.53

73.3

59.7

35.9

31.7

85.761

69.849

39.49

34.87


97.489

79.401

43.08

38.04

7.4 Site Curve and PAT curve Intersection Figure 7.1 PAT Curve with Maximum Diversion Factor and Minimum Diversion Factor

The intersecting point of the Pat curve and the Site curve (Site Curve is shown the Appendix 4) will give the operating point of the machine.

Figure 7.2 PAT Curves with intersecting with the Site Curve

7.5 Operating Point Operating point for 80 meter site:


Flow = 37 m^3/hr Head = 75 m Electrical Power output for 80 m site = 0.73*0.55*75*37*9.81*1000 = 3036 W Operating point for 60 meter site: Flow = 31 m^3/hr Head = 58 m Electrical Power output for 60 m site = 0.73*0.55*58*31*9.81*1000 = 1967 W Now consider the case when the head varies from 80 – 60 m. Since we have the PAT curve we can easily find the variation of the operating point of the PAT and also the variation in the output power.

Chapter 8 Cost-Benefit Analysis 8.1 Introduction A product is only helpful when it can come to the benefit of its users. But if it becomes impossible for a user to purchase the product, then the product comes to no use of the user. In the same way if the setup of the PAT generator system is very expensive, then people will not use it, rather they would search for some other alternatives. But the cost of this system is not that very high and a cost analysis will be discussed in this chapter. 8.2 Cost of the Setup What does the system is made up of? It has a Motor-pump, a reservoir, some pipes, valves and some capacitors. Cost of 3 phase 4 hp Motor-Pump

= BDT 50,000/-


Cost of Pipes per feet

= BDT 20/- per feet

Cost of Capacitor

= BDT 500/-

Cost of making a Reservoir

= BDT 100,000/-

Setup Cost

= BDT 5,000/-

We require 800 feet length pipe. So, Total cost of the setup is

= BDT (50,000+800*20+500+100,000+5,000) = BDT 171,500/-

8.3 Economic Benefits The maximum and minimum electrical power outputs from the system are 3075 W and 2212W respectively. Considering 2.4kW as generation can serve 24 houses if each house requires 1 CFL light (18 W), 1 fan (60 W) and 1 TV (40W). This setup will last for 10 years. Total Fixed Cost

= BDT 171,500/-

Total Fixed Cost per year

= BDT 17,150/-

Usually, very less maintenance is required for this kind of system. So, the maintenance cost is also very small. Total Maintenance cost is

= BDT 500/- per month

Annual Variable Cost

= BDT (500*12) = BDT 6,000/-

Considering the output power as 2.4kW. Capacity factor is 80%. 1 year is equivalent to 8760 hours Total Energy Generated per annum

= 2.4*8760*0.8 = 16820 kWhr


Total cost of generation per annum

= (Fixed Cost + Variable Cost)/Total Energy Generated = BDT (17,150+6,000)/16820 = BDT 1.38/- per kWhr

0 – 100 units of Energy cost is

= BDT 3.05/- per kWhr

Savings per kWhr is

= BDT (3.05 – 1.38)/= BDT 1.67/- per kWhr

Total savings per year

= BDT 1.67*16820 = BDT 28,089/-

Payback period

= Total Investment/Savings per year = 171,500/28,089 years = 6.1 years = 6 year 1.5 months.

It is very clear that the investment in this type of policy is also economically beneficial. The total invested amount is returned by 6 years 1.5 months. After that the electricity that is obtained is worth free. Moreover, National Grid power is usually not supplied in such hilly areas of Bangladesh. So, in a sense the power that the populace will receive is actually cost free.

Chapter 9 Discussion and Conclusion In this tough situation of our country PAT could be a probable solution of electricity where there is suitable sites. The people living in those areas could be benefited by the electricity in so many ways that is unimaginable. A small scale PAT electric Power Station can be set up across these pipes they use for the water flow. PAT electric power generation may be a potential energy source for urban areas in Bangladesh.


Our achievements through this thesis was only to understand how a motor-pump behaves if it is operated in the turbine mode. Our main work was to deal with the pump characteristic from where the turbine characteristic is found. Our study shows that we can certainly generate powers from this type of system. Furthermore the cost of setting up this system is less if we compare with any other sorts of power. We have also seen that the expense incurred in this setup is returned way before the lifetime of the setup is reached. Our study does not end here; we are hoping make a setup and observe the practical results.

Appendix A Conversion Factors for Head and Flow


Appendix B Turbine mode Performance away from the bep: Head vs Flow

Appendix B


Appendix C Turbine Mode Performance away from the bep: Power vs Flow


Appendix D Selecting Penstock Diameter and Calculating Head Loss


Hsite=80m Lpipe=233m α=20 degree Q(m^3/hr)

Q(l/s)

hf(%)

hf(m)

Hsite-hf(m)

5

1.388889

0.097269

0.077815

79.92218

10

2.777778

0.389075

0.31126

79.68874

15

4.166667

0.87542

0.700336

79.29966

20

5.555556

1.556302

1.245041

78.75496

25

6.944444

2.431721

1.945377

78.05462

30

8.333333

3.501678

2.801343

77.19866

35

9.722222

4.766173

3.812939

76.18706

40

11.11111

6.225206

4.980165

75.01984

Figure 1 Table for Head loss with head 80 m

Hsite=60m Lpipe=175m α=20 degree Q(m^3/hr)

Q(l/s)

hf(%)

hf(m)

Hsite-hf(m)

5

1.388889

0.097408

0.058445

59.94156

10

2.777778

0.389632

0.233779

59.76622

15

4.166667

0.876672

0.526003

59.474

20

5.555556

1.558528

0.935117

59.06488

25

6.944444

2.4352

1.46112

58.53888


30

8.333333

3.506688

2.104013

57.89599

35

9.722222

4.772992

2.863795

57.1362

40

11.11111

6.234112

3.740467

56.25953

Figure 2 Table for Head loss with head 80 m

References 9. 1. Electrical Power – Dr. S.L.Uppal 2. Renewable Energy Technologies – Chetan Singh Solanki 3. Motor As Generator for MicroHydro Power – Nigel 4. A Guide to Pump as Turbine Pico HydroPower Systems - Chris Greacen and Megan Kerins 5. The Performance of Centrifugal Pumps as Turbines and Influence of Pump Geometry – Arthur Williams and Arnaldo Rodrigues 6. Practical Centrifugal Pumps – Paresh Girdhar and Octo Moniz 7. Pumps As Turbines A user’s guide – Arthur Williams 8. Manual on Pumps Used as Turbines – J.-M. Chapallaz, P. Eichenberger, G. Fischer

Electric Machinery Fundamentals – Chapman

10. A Text Book of Electrical Technology – B L Thereja 11. http://en.wikipedia.org/wiki/Inducti

on_motor 12. http://www.google.com/imgres?

q=induction+motor 13. http://en.wikipedia.org/wiki/Turbin

e 14. http://www.koifishponds.com/new

_page_1.htm 15. http://www.pumpsasturbines.org.u

k 16. http://www.youtube.com/watch? v=bRl0ztxn094 17. http://www.pumpfundamentals.co

m/pumps_as_turbines.htm 18. http://www.pedrollo.co.uk


19. http://www.youtube.com/watch?

v=bRl0ztxn094


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